The average age of residents in a large residential retirement community is 69 years with standard deviation 5.8 years. A simple random sample of 100 residents is to be selected, and the sample mean age of these residents is to be computed.What is the standardised score (test statistic) for a particular sample of 100 resident that gave an average age, , equal to 70 years? Give three decimal places.

Question

BlackTom AI · Accepted Answer

1. Relevant knowledge points and concepts
The sampling distribution of the sample mean has mean equal to the population mean μ and standard deviation (standard error) equal to σ/√n when the population standard deviation σ is known. For large samples (or normal populations), the standardized test statistic (z-score) for a sample mean x̄ is z = (x̄ − μ) / (σ/√n).

2. Complete reasoning process step by step
We are given μ = 69, σ = 5.8, sample size n = 100, and observed sample mean x̄ = 70. Use the formula for the z-score of a sample mean because σ is known and n is large.

3. Calculations with all intermediate steps
Compute the standard error:
√n = √100 = 10
Standard error = σ/√n = 5.8 / 10 = 0.58

Compute the z-score:
Difference = x̄ − μ = 70 − 69 = 1
z = (x̄ − μ) / (σ/√n) = 1 / 0.58 = 1.724137931...

Round to three decimal places: z ≈ 1.724

4. Why the correct answer is correct
This z-value measures how many standard errors the observed sample mean (70) lies above the population mean (69). Because the population standard deviation is given and n = 100 is large, the sampling distribution approximation and the z-formula are appropriate, producing z = 1.724 (three decimals).

5. Why other common alternatives would be incorrect
- Using σ instead of σ/√n (z = 1/5.8 ≈ 0.172) ignores the reduction in variability when averaging many observations.
- Using σ/ n (i.e., dividing by 100) gives an absurdly large z (≈17.24) and misapplies the standard error formula.
- Using a t-statistic is unnecessary here because σ is known and n is large; if one mistakenly used t with small n, results would differ slightly but that is not relevant given the information.

6. Conclusion confirming the final answer
Therefore the standardized score (test statistic) is z = 1.724, which matches the provided answer.

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